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An Unsolvable Integral (according to Matlab)

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  1. May 18, 2005 #1
    Hello!

    I have the following problem: I cannot solve the integral below by the means of Matlab.
    [tex]
    \int_{-\infty}^{\infty} \frac{e^{-t^2}}{\left(2-t\right)^2 + 16} dt
    [/tex]​
    When I write the following in Matlab
    Code (Text):

    >> syms t;
    >> y = exp(-t^2) / (16 + (2 - t)^2);
    >> int(y, t, -inf, inf)
     
    it gives me the output :bugeye:
    Code (Text):

    Warning: Explicit integral could not be found.
    > In sym.int at 58
     
    ans =
     
    int(exp(-t^2)/(16+(2-t)^2),t = -Inf .. Inf)
     

    I managed to calculate the integral by the means of both Mathcad and Mathematica. Mathcad gave 0.088 as an answer (I had to explicitly specify "Infinite Limit" as a method). Mathematica gave me 0.0880741, I used the NIntegrate function:

    [tex]
    \mbox{NIntegrate}\left[\frac{e^{-t^2}}{\left(2-t\right)^2 + 16}, \left\{t, -\infty, \infty \right\} \right]
    [/tex]​

    Does anyone have an idea, how I can solve this integral in Matlab? What do Mathcad and Mathematica use in order to solve it?

    Thanks!
     
  2. jcsd
  3. May 18, 2005 #2

    dextercioby

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    I think it/they use(s) the theorem of residues.The integrand has simple poles at [itex] 2\mp 4i [/itex].

    Daniel.
     
  4. May 18, 2005 #3

    dextercioby

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    I didn't find this integral in G & R 5-th edition,CD version

    [tex] \int_{0}^{\infty} \frac{e^{-ax^{2}+bx+c}}{x^{2}+d^{2}} \ dx [/tex]

    ,but this one was

    [tex] \int_{0}^{\infty} \frac{e^{-\mu^{2}x^{2}}}{x^{2}+b^{2}} \ dx [/tex]

    Daniel.
     
    Last edited: May 18, 2005
  5. May 19, 2005 #4
    Could you please explain what "G & R" is?
     
  6. May 19, 2005 #5

    dextercioby

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    Gradshtyn & Rytzhik,"Tables of Series,Integrals and Products",Academic Press,5-th edition,CD version.

    Daniel.
     
  7. May 19, 2005 #6

    dextercioby

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    Here it is,courtesy of Mathematica,a closely related integral.

    [tex] \int_{-\infty}^{+\infty} \frac{e^{-x^{2}}}{(2-x)^{2}+4} \ dx =\frac{\sqrt{\pi}}{6}\left[3\sqrt{\pi}\cos 8-12 \ _{1}F_{2}\left(1,\frac{3}{4},\frac{5}{4};-16\right) +64 \ _{1}F_{2}\left(1,\frac{5}{4},\frac{7}{4};-16\right)\right] [/tex]

    Daniel.
     
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